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Q23:
Find all points of discontinuity of $f$, where $f(x) = \begin{cases} \frac{\sin x}{x}, & \text{if } x < 0 \\ x + 1, & \text{if } x \ge 0 \end{cases}$

Solution :

Step 1: Analyzing the Domain of Continuity

The function $f(x)$ is defined piecewise over the set of real numbers $\mathbb{R}$. We must test for continuity at the point of interest, $x = 0$, where the definition of the function transitions, and evaluate the function behavior in the open intervals $(-\infty, 0)$ and $(0, \infty)$.

Step 2: Analysis at $x < 0$

For $x \in (-\infty, 0)$, $f(x) = \frac{\sin x}{x}$.
Since $g(x) = \sin x$ and $h(x) = x$ are continuous functions on their entire domains, their quotient $\frac{g(x)}{h(x)}$ is continuous provided the denominator $h(x) \neq 0$ [Per the Quotient Rule for Continuous Functions]. Since $x \neq 0$ in this interval, $f(x)$ is continuous for all $x < 0$.

Step 3: Analysis at $x > 0$

For $x \in (0, \infty)$, $f(x) = x + 1$.
This is a polynomial function of degree 1. Polynomials are continuous for all $x \in \mathbb{R}$ [Per the Polynomial Continuity Theorem]. Thus, $f(x)$ is continuous for all $x > 0$.

Step 4: Analysis at the Critical Point $x = 0$

To determine if $f(x)$ is continuous at $x = 0$, we must satisfy the condition $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.

  • Left-Hand Limit (LHL): $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin x}{x}$. Applying the standard trigonometric limit, $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ [By L'Hôpital's Rule or Small Angle Approximation Taylor Series], the LHL = 1.
  • Right-Hand Limit (RHL): $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x + 1) = 0 + 1 = 1$.
  • Function Value at $x = 0$: $f(0) = 0 + 1 = 1$ (using the definition for $x \ge 0$).

Since $\text{LHL} = \text{RHL} = f(0) = 1$, the function $f(x)$ is continuous at $x = 0$.

Step 5: Visual Representation

0 x y

Summary Analysis

Interval Continuity Status
$x < 0$ Continuous
$x = 0$ Continuous
$x > 0$ Continuous

Final Solution: The function $f(x)$ is continuous for all $x \in \mathbb{R}$. Therefore, there are no points of discontinuity.


More Questions from Class 12 Mathematics Continuity and Differentiability EXERCISE 5.1


CBSE Solutions for Class 12 Mathematics Continuity and Differentiability


Chapters in CBSE - Class 12 Mathematics


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